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If $m$ is a positive integer,then how can I show that $\int_{0}^{\frac{\pi}{2}} \cos^m x\sin^m x dx=2^{-m}\int_{0}^{\frac{\pi}{2}} \cos^m x dx$?

I tried substitution but to no avail.I think I am going horribly wrong.

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2 Answers 2

up vote 3 down vote accepted

Well, $2\sin(x)\cos(x)=\sin(2x)$, so

$$\int_0^{\pi/2}\cos^m(x)\sin^m(x)dx=\int_0^{\pi/2}\left(\frac{1}{2}\sin(2x)\right)^mdx=\frac{1}{2^m}\int_0^{\pi/2}\sin^m(2x)dx$$

From there, it should be quite a bit easier to handle.

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Once you have reduced your problem to finding $\int_0^{\pi/2}\sin^{2m}(2x)\,dx$, as described by Jack Maney, the substitution $u=2x$ is almost automatic, and we find that $$\int_0^{\pi/2}\sin^{2m}(2x)\,dx=\frac{1}{2}\int_0^{\pi} \sin^m u\,du.$$ Now we use symmetry. The curve $y=\sin^{m}u$ is symmetrical about $u=\pi/2$, so the area under $y=\sin^m u$, above the $u$-axis, from $u=0$ to $u=\pi$ is twice the area from $0$ to $\pi/2$.

And the area under $y=\sin^m u$ from $0$ to $\pi/2$ is the same as the area under $y=\cos^m u$ from $0$ to $\pi/2$. This is clear from a picture, or else we can make the substitution $v=\pi/2-u$.

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